Proof that closure of a space equals another space.

1. Nov 29, 2012

Paalfaal

1. The problem statement, all variables and given/known data

Define:

- c0 = {(xn)n $\in$ $\ell$$\infty$ : limn → $\infty$ xn = 0}

- l0 = {(xn)n $\in$ $\ell$$\infty$ : $\exists$ N $\in$ the natural numbers, (xn)n = 0, n $\geq$ N}

Problem: Prove that $\overline{\ell}$0= c0 in $\ell$$\infty$

2. Relevant equations

3. The attempt at a solution

I want to find the solution using the limit-definition of closure.

Considering an element

x = (xn) $\in$ c0

and a sequence

yj = (xjn) $\in$ $\ell$0,

such that xnj = xn for n < j, xnj = 0, otherwise.

Using the metric induced by the supremum norm; || xn - xnj || $\rightarrow$ 0 as j tends to infinity. We can du this for all elements in c0, and hence c0 $\subseteq$ $\overline{\ell}$0.

My problem is to show the other direction, that is

$\overline{\ell}$0 $\subseteq$ c0

I need to show that elements in $\overline{\ell}$0 has a vanishing limit. I don't know how to do this using the supremum norm. In fact, it seems impossible to me..

Can I get any help?

2. Nov 29, 2012

micromass

Staff Emeritus
Clearly, we know that $\ell_0\subseteq c_0$. Try to prove now that $c_0$ is closed. This would imply what you try to prove.

3. Nov 29, 2012

Paalfaal

Hmmm.. When is it true that $A \subseteq B \Rightarrow$ $\bar{A}$ $\subseteq$ $\bar{B}$ ?

I ask because I've been working on a similar problem: Prove that $\bar{\ell}$0 = $\ell$2 in $\ell$2.

With a similar approach of the attempt above I got $\bar{\ell}$0 $\supseteq$ $\ell$2.

Since $\ell$0 $\subseteq$ $\ell$2 (for obvious reasons),
$\bar{\ell}$0 $\subseteq$$\ell$2 iff $\ell$2 = $\bar{\ell}$2 (correct me if I'm wrong here).

In that case, $\bar{\ell}$0 = $\ell$2 in $\ell$2.

(I guess it might is trivial that $\ell$2 = $\bar{\ell}$2 in $\ell$2, but Is there an easy way to prove this?)

4. Nov 29, 2012

micromass

Staff Emeritus
This is always true. Try to prove it.

If X is a metric space, then $\overline{X}=X$ (closure in X). Try to prove this.

5. Nov 29, 2012

Paalfaal

I certiantly will!

Thank you for your help! Much appreciated